Answer:
The answer to the question is
The solution contains 4.73 % by mass of NaCl
Explanation:
To solve this we list out the variables thus
The Molarity of NaCl = 0.850 M,
Density of solution = 1.05 g/mL
Molar mass of NaCl = 58.44 g/mol
Therefore mass of NaCl in he solution = (58.44 g/mol) × (0.850 M) = 49.674 g of NaCl
However the density = 1.05 g/mL or 1050 g/L
Therefore the mass percent of NaCl in 1 liter of solution or 1050 g of solution = (49.674/1050) × 100 = 4.73 %
The concentration of NaCl in mass percent = 4.73 %
Mg( s ) | Mg 2+ ( aq ) || Al 3+ ( aq ) | Al( s )
It's likely what's wanted is
<span><span>Li</span>→<span><span>Li</span><span>2+</span></span>+2<span>e−
</span></span>
The reason is because IEs are usually reported from the neutral atom, that is, IE2 is the energy required to remove two electrons from a neutral Li atom, as above, rather than the additional energy required to remove one more electron from an Li+ cation.
The solution for this problem is:
(1.55x10^4 / 1.0x10^3) x 19.8 mm Hg
= 15.5 x 19.88 mm Hg
= 308.14 mm Hg decrease
= 308.14 x 0.05 C = 15.407 deg C
deduct this amount to 100
100 – 15.407 = 84.593 C
ANSWER: 85 deg C (rounded to 2 significant figures)
